Evaluate $\int \frac{1}{x \sqrt{16x^2-9}}dx$

Question

I need help with this exercise.

Evaluate $\int \frac{1}{x \sqrt{16x^2-9}}dx$

This is why I've tried but it seems wrong.

$$\int \frac{1}{x \sqrt{16x^2-9}}dx$$

$$=\int \frac{1}{x \sqrt{(4x)^2-3^2}}dx$$

$$=\int \frac{4 \cdot 1}{4 \cdot x \sqrt{(4x)^2-3^2}}dx$$

$$=4\int \frac{1}{4x \sqrt{(4x)^2-3^2}}dx$$

$$=4\left(\frac{1}{3} \sec^{-1} \frac{|4x|}{3}\right) + C$$

I've checked in multiple math solvers that gives me the answer but not the process and answer is the same but without the 4 that is in the beginning. Is there a way that I can do it without trigonometric substitution? I know that this can be done with trigonometric substitution (I've already tried it and got the answer right) but this exercise appears in a section of a book of derivatives and integrals of Inverse Trigonometric Functions which is before the chapter where all the techniques of integrations appears. So this is supposed to be done without using that knowledge. So I'm wondering how I can get the answer using a simpler way.

Thanks for your help.

Answer 1

I found the "simpler way". :-S

Is using substitution with $u=4x$, so $x=\frac{u}{4}$, then $du=4dx$, so $\frac{du}{4}=dx$.

$$\int \frac{1}{x \sqrt{16x^2-9}}dx$$

$$=\int \frac{1}{x \sqrt{(4x)^2-3^2}}dx$$

$$=\int \frac{1}{\frac{u}{4} \sqrt{u^2-3^2}}\cdot\frac{du}{4}$$

$$=\int \frac{1}{u\sqrt{u^2-3^2}}du$$

$$=\frac{1}{3}sec^{-1}\frac{|u|}{3}+C$$

$$=\frac{1}{3}sec^{-1}\frac{|4x|}{3}+C$$

Answer 2

In the last step, you're forgetting to use the chain rule: $$\int \frac{1}{4x \sqrt{(4x)^2-3^2}}dx=\frac14\int \frac{4}{4x \sqrt{(4x)^2-3^2}}dx=\frac{1}{4}\frac 13 \sec^{-1}\frac{|4x|}{3}+c$$ Hence: $$4\int \frac{1}{4x \sqrt{(4x)^2-3^2}}dx= \frac 13 \sec^{-1}\frac{|4x|}{3}+c$$

Answer 3

I have seen this taught in such a way that we find the derivatives of the inverse trig functions.

$\frac {d}{dx} \sec^{-1} x = \frac {1}{x\sqrt{x^2 - 1}}\\ \frac {d}{dx} \sec^{-1} ax = \frac {1}{x\sqrt{a^2x^2-1}}\\ \frac {d}{dx} \frac 1a \sec^{-1} \frac {x}{a} = \frac {1}{x\sqrt{x^2 - a^2}}\\ \frac {d}{dx} \tan^{-1} x = \frac {1}{x^2 + 1}$

etc.

So, rather than using the substitution $x = \frac 34 \sec t$ and simplifying. We would try to make the integral into a constant times something we can find on a table of known derivatives.

$\int \frac {1}{x\sqrt{16x^2 - 9}}\ dx\\ \int \frac {1}{3x\sqrt{(\frac 43x)^2 - 1}}\ dx$

Which is on our table.

This is far less versitile than the substitution

$x = \frac 34 \sec t\\ dx = \frac 34 \sec t\tan t \ dt$

$\int \frac { \frac 34 \sec t\tan t \ dt}{\frac 34\sec t\sqrt {9\tan^2 t}} \ dt\\ \frac 13 \int \ dt\\ \frac 13 t + C$

and reverse the substitution.

$t = \sec^{-1} \frac 43x\\ \frac 13 \sec^{-1} \frac 43x + C$

Answer 4

$$ \begin{aligned} \int \frac{1}{x \sqrt{16 x^2-9}} d x = & \frac{1}{16} \int \frac{1}{x^2} d\left(\sqrt{16 x^2-9}\right) \\ = & \int \frac{d\left(\sqrt{16 x^2-9}\right)}{\left(\sqrt{16 x^2-9}\right)^2+9} \\ = & \frac{1}{3} \tan ^{-1}\left(\frac{\sqrt{16 x^2-9}}{3}\right)+C \end{aligned} $$

Answer 5

This answer is the same as the answer with secant inverse, since $\arcsecx + \arccosecx = \pi/2$

Answer 6

HINT

WLOG, let us assume that $x > 3/4$. I would recommend you to notice that \begin{align*} \int\frac{\mathrm{d}x}{x\sqrt{16x^{2} - 9}} & = \int\frac{\mathrm{d}x}{x^{2}\sqrt{16 - 9x^{-2}}}\\\\ & = -\int\frac{\mathrm{d}(1/x)}{\sqrt{16 - 9x^{-2}}}\\\\ & = -\int\frac{\mathrm{d}u}{\sqrt{16 - 9u^{2}}} \end{align*} Which is (almost) a standard integral.

Can you take it from here?